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Question: If \(\left| x \right|<1\), then the sum of the series \(1+2x+3{{x}^{2}}+4{{x}^{3}}+.....\infty \) wi...

If ∣x∣<1\left| x \right|<1, then the sum of the series 1+2x+3x2+4x3+.....∞1+2x+3{{x}^{2}}+4{{x}^{3}}+.....\infty will be
A. 11−x\dfrac{1}{1-x}
B. 11+x\dfrac{1}{1+x}
C. −11+x2-\dfrac{1}{1+{{x}^{2}}}
D. 1(1−x)2\dfrac{1}{{{\left( 1-x \right)}^{2}}}

Explanation

Solution

We use (1−x)n=1−nx+n(n−1)2!x2−......+(−1)rn(n−1)....(n−r+1)r!xr+....+∞{{\left( 1-x \right)}^{n}}=1-nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}-......+{{\left( -1 \right)}^{r}}\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty , the infinite binomial formula. We replace the value of n=−2n=-2. We simplify the coefficients and get the simplest forms. We convert the indices to get the required solution.

Complete answer:
We know that the binomial theorem of (1+x)n,∣x∣<1,n∈R{{\left( 1+x \right)}^{n}},\left| x \right|<1,n\in \mathbb{R} gives
(1+x)n=1+nx+n(n−1)2!x2+......+n(n−1)....(n−r+1)r!xr+....+∞{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+......+\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty .
Now when the value of nn is a negative real number we get
(1+x)n=1+nx+n(n−1)2!x2+......+n(n−1)....(n−r+1)r!xr+....+∞{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+......+\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty
We first replace the value of xx with −x-x to get
(1−x)n=1−nx+n(n−1)2!x2−......+(−1)rn(n−1)....(n−r+1)r!xr+....+∞{{\left( 1-x \right)}^{n}}=1-nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}-......+{{\left( -1 \right)}^{r}}\dfrac{n\left( n-1 \right)....\left( n-r+1 \right)}{r!}{{x}^{r}}+....+\infty .
We now put the value for n=−2n=-2 to get
(1−x)−2=1−(−2)x+(−2)(−2−1)2!x2−(−2)(−2−1)(−2−2)3!x3+.......∞{{\left( 1-x \right)}^{-2}}=1-\left( -2 \right)x+\dfrac{\left( -2 \right)\left( -2-1 \right)}{2!}{{x}^{2}}-\dfrac{\left( -2 \right)\left( -2-1 \right)\left( -2-2 \right)}{3!}{{x}^{3}}+.......\infty .
We now simplify to get

& {{\left( 1-x \right)}^{-2}} \\\ & =1-\left( -2 \right)x+\dfrac{\left( -2 \right)\left( -2-1 \right)}{2!}{{x}^{2}}-\dfrac{\left( -2 \right)\left( -2-1 \right)\left( -2-2 \right)}{3!}{{x}^{3}}+.......\infty \\\ & =1+2x+3{{x}^{2}}+4{{x}^{3}}+.....\infty \\\ \end{aligned}$$ Therefore, the required sum is $${{\left( 1-x \right)}^{-2}}=\dfrac{1}{{{\left( 1-x \right)}^{2}}}$$. **And hence the correct answer is option D.** **Note:** An infinite series has a sum that is convergent in the limit sense. The limiting value can be found from $${{\left( 1+x \right)}^{n}}$$. In the case of a fraction value of $n$ we can also use the infinite form. It can take both negative and positive but for an integer, it has to be negative.